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Proof that closure of a space equals another space.

  1. Nov 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Define:

    - c0 = {(xn)n [itex]\in[/itex] [itex]\ell[/itex][itex]\infty[/itex] : limn → [itex]\infty[/itex] xn = 0}

    - l0 = {(xn)n [itex]\in[/itex] [itex]\ell[/itex][itex]\infty[/itex] : [itex]\exists[/itex] N [itex]\in[/itex] the natural numbers, (xn)n = 0, n [itex]\geq[/itex] N}


    Problem: Prove that [itex]\overline{\ell}[/itex]0= c0 in [itex]\ell[/itex][itex]\infty[/itex]

    2. Relevant equations


    3. The attempt at a solution

    I want to find the solution using the limit-definition of closure.

    Considering an element

    x = (xn) [itex]\in[/itex] c0

    and a sequence

    yj = (xjn) [itex]\in[/itex] [itex]\ell[/itex]0,

    such that xnj = xn for n < j, xnj = 0, otherwise.

    Using the metric induced by the supremum norm; || xn - xnj || [itex]\rightarrow[/itex] 0 as j tends to infinity. We can du this for all elements in c0, and hence c0 [itex]\subseteq[/itex] [itex]\overline{\ell}[/itex]0.

    My problem is to show the other direction, that is

    [itex]\overline{\ell}[/itex]0 [itex]\subseteq[/itex] c0

    I need to show that elements in [itex]\overline{\ell}[/itex]0 has a vanishing limit. I don't know how to do this using the supremum norm. In fact, it seems impossible to me..

    Can I get any help?
     
  2. jcsd
  3. Nov 29, 2012 #2

    micromass

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    Clearly, we know that [itex]\ell_0\subseteq c_0[/itex]. Try to prove now that [itex]c_0[/itex] is closed. This would imply what you try to prove.
     
  4. Nov 29, 2012 #3
    Hmmm.. When is it true that [itex]A \subseteq B \Rightarrow[/itex] [itex]\bar{A}[/itex] [itex]\subseteq[/itex] [itex]\bar{B}[/itex] ?


    I ask because I've been working on a similar problem: Prove that [itex]\bar{\ell}[/itex]0 = [itex]\ell[/itex]2 in [itex]\ell[/itex]2.

    With a similar approach of the attempt above I got [itex]\bar{\ell}[/itex]0 [itex]\supseteq[/itex] [itex]\ell[/itex]2.

    Since [itex]\ell[/itex]0 [itex]\subseteq[/itex] [itex]\ell[/itex]2 (for obvious reasons),
    [itex]\bar{\ell}[/itex]0 [itex]\subseteq[/itex][itex]\ell[/itex]2 iff [itex]\ell[/itex]2 = [itex]\bar{\ell}[/itex]2 (correct me if I'm wrong here).

    In that case, [itex]\bar{\ell}[/itex]0 = [itex]\ell[/itex]2 in [itex]\ell[/itex]2.


    (I guess it might is trivial that [itex]\ell[/itex]2 = [itex]\bar{\ell}[/itex]2 in [itex]\ell[/itex]2, but Is there an easy way to prove this?)

    Any comments?

    :smile:
     
  5. Nov 29, 2012 #4

    micromass

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    This is always true. Try to prove it.

    If X is a metric space, then [itex]\overline{X}=X[/itex] (closure in X). Try to prove this.
     
  6. Nov 29, 2012 #5
    I certiantly will!


    Thank you for your help! Much appreciated :smile:
     
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